The reaction cis- `X underset(k_(b))⇌overset(k_(f))` trans-X is first order in both directions. At `25^(@)` C, the equilibrium constant is 0.10 and the rate constant `k_(f) = 3 xx 10^(-4) s^(-1)` . In an experiment starting with the pure cis-form, how long would it take for half of the equilibrium amount of the trans-isomer to be formed?

Options

A. 150 sec
B. 200sec
C. 240 sec
D. 210 sec

Explanation

cis `- Xunderset(k_(b))overset(k_(f))hArr` trans - X
`{:("Initial",a,0,),("at time t",a-x,x,),("at eqm",a-x_(e),x_(e),):}`
`K_((eq)) = (k_(e))/(k_(b)), k_(b) = (3 xx 10^(-4))/(0.1) = 3 xx 10^(-3)` As we know `(k_(f) + k_(b)) = (1)/(t) ln((x_(e))/(x_(e)x))`
given `x = (x_(e))/(2) " "(k_(f) + k_(b)) = (1)/(t) ln 2` or `(3 xx 10^(-3) + 3 xx 10^(-4)) = (0.693)/(t) " " t = 210 sec`